Question

Suppose that \(f^{\prime}(0)\) exists and \(f(x+y)=f(x) f(y)\) for all \(x\) and \(y\). Prove that \(f^{\prime}\) exists for all \(x\).

Short answer

Question: Prove that if a function \(f(x)\) exists such that \(f'(0)\) exists and \(f(x+y) = f(x)f(y)\) for all \(x\) and \(y\), then \(f'(x)\) must exist for all \(x\). Answer: The function \(f'(x)\) exists for all \(x\) as it can be represented as the product of two existing functions, specifically, \(f(x) f'(0)\).

Step-by-step solution

Utilizing the given expression

First, we will use the given expression \(f(x+y) = f(x)f(y)\) and manipulate it to study the behavior of the function. When \(y = x\), we have \(f(2x) = f(x)^2\).

Manipulate the expression for f(x)

Now, divide both sides by \(f(x)^2\) to get \(\frac{f(2x)}{f(x)^2} = 1\). Then, rewrite the expression in terms of \(h = x\). So our new expression is \(\frac{f(2h)}{f(h)^2} = 1\), which will be important in the next step.

Use the definition of derivative to find \(f'(x)\)

To show that \(f'(x)\) exists, we will use the definition of derivative: \(f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\). Now, using our given expression \(f(x+y) = f(x)f(y)\), we can rewrite this as \(f'(x) = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h}\).

Finding the derivative at x = 0

We know that \(f'(0)\) exists, so let's find it: \(f'(0) = \lim_{h \to 0} \frac{f(h) - 1}{h}\)

Prove that the derivative exists for all x

Notice that our expression for \(f'(x)\) in step 3 is dependent on the limit that we now know exists at \(x = 0\): \(f'(x) = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h} = f(x) f'(0)\) Since we know that \(f'(0)\) exists, and \(f(x)\) is a function, we can conclude that the derivative exists for all \(x\) as the product of two existing functions. Therefore, \(f'(x)\) exists for all \(x\).

Key concepts

Functional Equation

A functional equation is a mathematical expression that asserts the equality of two functions for all arguments they are defined for. In our exercise, the functional equation given is \(f(x+y) = f(x)f(y)\). This type of equation helps to define or constrain the nature of a function and often plays a central role in problems related to functions in mathematics.

The significance of a functional equation lies in its ability to establish a relationship between the values of a function at different points. For instance, from the equation \(f(x+y) = f(x)f(y)\), by setting \(y=0\), we get \(f(x) = f(x)f(0)\), which implies if \(f(0) \eq 0\), then \(f(x)\) must be equal to 1 for all \(x\), given that \(f(0)\) is the multiplicative identity for the function. Understanding functional equations is vital for solving complex problems in real analysis and other areas of mathematics.

Definition of Derivative

The definition of derivative is at the heart of calculus. It describes how a function changes as its input changes. Formally, the derivative of a function \(f\) at a point \(x\) is defined by the limit \( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \).

Intuitively, the derivative represents the slope or rate of change of the function \(f\) at the point \(x\). If this limit exists, we say that \(f\) is differentiable at \(x\), and \(f'(x)\) represents the instantaneous rate of change. In our exercise, we are given that this derivative exists at \(x=0\), which is a critical piece of information for proving that the derivative exists at all points. The step-by-step solution uses the definition of the derivative along with the given functional equation to establish that \(f'(x)\) exists for all \(x\) by connecting it with the known derivative at \(x=0\).

Limit of a Function

The limit of a function is a fundamental concept in calculus, which describes the behavior of a function as its argument approaches a certain value. In mathematical terms, \( \lim_{x \to a} f(x) \) signifies the value that the function \(f\) approaches as \(x\) gets closer and closer to \(a\).

If this limit exists and is equal to some real number \(L\), we can say that the function \(f\) approaches the value \(L\) as \(x\) approaches \(a\). This concept becomes crucial when dealing with derivatives, as the definition of a derivative itself is based on the idea of a limit. In the problem we're discussing, the derivative of the function \(f\) at any point \(x\) is computed by finding the limit of the difference quotient as \(h\) approaches zero. The existence of the limit as \(h \to 0\) is what allows us to conclude that \(f'(x)\) exists for all \(x\).

Real Analysis

Real analysis is a branch of mathematics that deals with the set of real numbers and the functions defined on them. This field encompasses several key concepts, including limits, continuity, differentiation, integration, and sequences and series. Real analysis provides a rigorous underpinning for calculus, and it is where many of the intuitive ideas from calculus are made precise.

In real analysis, we often deal with proving the existence of limits and derivatives, the continuity of functions, and the convergence of sequences and series. The exercise we are looking at falls squarely within the realm of real analysis, as it challenges us to demonstrate the existence of derivatives across all points for a particular function defined by a given functional equation. Understanding these concepts deeply strengthens a student's mathematical foundation and provides the tools necessary to tackle advanced problems in mathematics and applied sciences.